MATH3201 2016-2017 Homework 13&14 - Solutions PDF

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Dr. Anna Felikson, Durham University Geometry, 21.2 Solutions 13-14 13. Draw in each of the two conformal models (Poincare disc and upper half-plane): (a) two intersecting lines; (b) two parallel lines; (c) two ultra-parallel lines; (d) infinitely many disjoint (hyperbolic) half-planes; (e) a circle...

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Dr. Anna Felikson, Durham University

Geometry, 21.2.2017

Solutions 13-14 13.1. Draw in each of the two conformal models (Poincare disc and upper half-plane): (a) two intersecting lines; (b) two parallel lines; (c) two ultra-parallel lines; (d) infinitely many disjoint (hyperbolic) half-planes; (e) a circle tangent to a line. Solution: One solution is indicated by colors: (a), (b), (c), (d), (e).

13.2. In the upper half-plane model draw (a) a (hyperbolic) line through the points i and i + 2; (b) a (hyp.) line through i+1 orthogonal to the (hyp.) line represented by the ray {ki | k > 0}; (c) a (hyperbolic) circle centred at i (just sketch it, no formula needed!); (d) a triangle with all three vertices at the absolute (such a triangle is called ideal). Solution: ki

i

i+2

i+1

i

13.3. Prove SSS, ASA and SAS theorems of congruence of hyperbolic triangles. Solution: 1. SSS: Let A1 B1 C1 and A2 B2 C2 be two hyperbolic triangles satisfying SSS (i.e. having the same side lengths). First we apply an isometry which takes A2 to A1 and B2 to B1 . Then the points C1 and C2 lie on the intersection of (hyperbolic) circles γA (centred at A1 of radius A1 C1 ) and γB (centred at B1 of radius B1 C1 ). Consider these circles in Poincar´e disc or half-plane model. Since hyperbolic circles are represented by Euclidean circles, two circles have at most two intersection points. Moreover, the two intersection points are symmetric with respect to the (hyperbolic) line A1 B1 Hence, there is an isometry which takes A2 B2 C2 to A1 B1 C1 . 2. ASA: Suppose that A1 B1 = A2 B2 ∠A1 = ∠A2 and ∠B1 = ∠B2 . Apply an isometry which takes s A2 to A1 and B2 to B1 . Then C1 and C2 lie on a (hyperbolic) ray starting from A1 1

and making angle ∠A1 with A1 B1 . There exists exactly one such a ray in each half-plane with respect to the line A1 B1 (this is especially clear if A1 is the centre of the Poincar´e disc model). Similarly, C1 and C2 lie on a (hyperbolic) ray starting from B1 and making angle ∠B1 with A1 B1 . As two rays have at most one intersection, we get at most one possibility for the point C1 and C2 in each of two half-planes, also these two possibilities are symmetric with respect to A1 B1 . Hence, A2 B2 C2 may be transformed to A1 B1 C1 by an isometry. 3. SAS: Suppose that A1 B1 = A2 B2 , ∠A1 = ∠A2 and A1 C1 = A2 C2 . First, map the angle ∠A2 to ∠A1 , then the points C1 and C2 lie on the given distances on the given lines. 13.4. Let ABC be a triangle. Let B1 ∈ AB and C1 ∈ AC be two points such that ∠AB1 C1 = ∠ABC . Show that ∠AC1 B1 > ∠AC B . Solution: Consider the quadrilateral B1 BC C1 . If ∠AC1 B1 ≤ ∠ACB then the sum of angles of B1 BC C1 is greater or equal to 2π. On the other hand, we can divide B1 BC C1 by a diagonal into two triangles, each having a sum of angles less than π. The contradiction proves ∠AC1 B1 > ∠ACB. 13.5. Show that there is no “rectangle” in hyperbolic geometry (i.e. no quadrilateral has four right angles). Solution: Suppose there is a rectangle ABC D. Then its sum of angles is 2π. Decompose it into two triangles by a diagonal AC. The sum of angles of ABC is smaller than π, the sum of angles of ACD is smaller than π but the sum of these two sums of angles equal to the sum of angles of ABCD. Contradiction. 13.6. (*) Given an acute-angled polygon P (i.e. a polygon with all angles smaller or equal to π/2) and lines m and l containing two disjoint sides of P , show that l and m are ultra-parallel. Solution: Let A1 A2 . . . An be an acute-angled n-gon. First, let as prove that the lines containing two “almost adjacent” rays A1 A2 and A4 A3 are disjoint: i.e. suppose B = A1 A2 ∩ A4 A3 . Then the triangle A2 A3 B has at least two non-acute angles, which contradicts to the fact the angle sum of a hyperbolic triangle is less than pi. Similarly, assuming that the rays A1 A2 and A5 A4 do intersect, we see a (non-convex) quadrilateral, with two non-acute angles and one angle bigger than π, which again contradicts the angle sum. In general, intersection of the rays A1 A2 and Ak Ak−1 will result in an (k − 1)-gon breaking the angle sum theorem. 14.7. Given α, β, γ such that α + β + γ < π, show that there exists a hyperbolic triangle with angles α, β, γ. Solution: Put a vertex A of angle α in the centre of the Poincar´e disc model. Let AX and AY be the rays forming the angle. Let T be a point on AX. Let T Z be a ray emanating from T and such that ∠ZT A = β. Consider the point C(T ) = T Z ∩ AY . When T is very close to A the hyperbolic line T Z is very close to a Euclidean line through the same points, so the hyperbolic sum of angles of triangle AT C(T ) is very close to π. As T runs away from A to Y , angle AC(T )T became smaller and smaller (see Question 13.4), then it tends to zero (when T Z is parallel to AY ) and disappears. By the continuity we see that for some intermediate point T0 the sum of angles of AT0 C(T0 ) coincides with α + β + γ which implies that ∠AC(T0 )T0 = γ and AT0 C(T0 ) is the required triangle. 14.8. Show that there exists a hyperbolic pentagon with five right angles. Solution: Consider a Euclidean regular pentagon PEucl . Draw PEucl so that the centre of PEucl coincides with the centre O of Poincar´e disc model. Consider the hyperbolic pentagon P spanned by the vertices of PEucl . When PEucl is very small (but still centred at O) the angle of P are almost the same as the angles of PEucl . When PEucl is inscribed into the absolute the angles of P are zero (as the adjacent sides of P are tangent). Notice that the angles of PEucl 2

are obtuse (more precisely, they are equal to 3π/5). So, by continuity we see that there exist some intermediate size of PEucl such that the corresponding hyperbolic regular pentagon has right angles. 14.9. (*) An ideal triangle is a hyperbolic triangle with all three vertices an the absolute. (a) Show that all ideal triangles are congruent. (b) Show that the altitudes of an ideal triangle are concurrent. (c) Show that an ideal triangle has an inscribed circle. Solution: (a) There exists a hyperbolic isometry which takes any given triple of points on the absolute to any other triple. So, it takes a triangle spanned by the given three points to the triangle spanned by the other three points. (b) Part (a) implies that any ideal triangle may be represented by a “regular” ideal triangle in the Poincar´e disc model (i.e. by a triangle with vertices 1, eiπ/3 , e2iπ/3 ). The symmetry shows that all altitudes of this triangle pass through the centre O of the model, so, are concurrent. (c) Similarly to part (b), looking at the “regular” representative, we can see that there is an inscribed circle centred at O (we can take a very small circle and start to blow it up till it will touch one of the sides; by symmetry reasons it will touch all other sides at the same time). 14.10 (*) We have proved that an isometry fixing 3 points of the absolute is identity map. How many isometries fix two points of the absolute? Solution: We will work in the upper half-plane model. Let f be an isometry fixing two points of the absolute. First, we can conjugate f by an isometry h which takes the fixpoints of f to 0 and ∞. Then h−1 ◦ f ◦ h fixes 0 and ∞. Moreover, for every isometry f ′ fixing the same two points as f , the isometry h−1 ◦ f ′ ◦ h fixes 0 and ∞. This implies that for answering the question it is sufficient to consider isometries fixing 0 and ∞. Now, any orientation-preserving isometry of the upper half-plane may be written as az+b with cz+d real a, b, c, d. Preserving 0 and ∞ means b = 0 and c = 0, so any orientation-preserving isometry fixing 0 and ∞ writes as az, a ∈ R+ . Hence, we get 1-parametric family of orientation-preserving isometries (all hyperbolic). Similarly, we obtain a one-parametric family of orientation-reversing isometries −a¯ z , a ∈ R+ . 14.11 (a) Show that the group of isometries of hyperbolic plane is generated by reflections. (b) How many reflections do you need to map a triangle ABC to a congruent triangle A′ B ′ C ′ ? Solution: We can do (a) and (b) simultaneously, using the same procedure as in E2 or S 2 : first apply reflection r1 to take A to A′ (with respect to the perpendicular bisector); then use the reflection in A′ M (where M is a midpoint of A′ r(A)); last, if needed, the reflection in A′ B ′ . So, we need at most 3 reflections. 14.12 (*) (a) Does there exist a regular triangle in hyperbolic plane? (b) Does there exist a right-angled regular polygon in hyperbolic plane? How many edges does it have (if exists)? Solution: (a) In the Poincar´e disc model, consider a triangle whose vertices are represented by vertices of a regular Euclidean triangle with centre at O. By symmetry, this triangle is also a regular hyperbolic triangle (in other words, all Euclidean isometries we would use to check that the Euclidean triangle is regular are also isometries of the hyperbolic plane). 3

(b) The same construction as in (a) shows that there is a regular n-gon for all integer n ≥ 3 in hyperbolic plane. When we make this n-gon very small, its sides are almost Euclidean lines, so its angles are almost the same as the angles of a regular Euclidean n-gon, i.e. (n−2)π/n. When the regular Euclidean n-gon grows, the angles of the corresponding hyperbolic n-gon decrease monotonically (to see this use Question 13.4), when all vertices of the n-gon are on the absolute, the angles are 0. So, the angles take every intermediate value between (n − 2)π/n and 0. In particular, if n > 4 then (n − 2)π/n > π/2, which implies that there is a right-angled n-gon for every n > 4. We also know (from the sum of angles) that there are no right angled triangles and quadrilaterals. 14.13 (a) Show that the angle bisectors in a hyperbolic triangle are concurrent. (b) Show that every hyperbolic triangle has an inscribed circle. (c) Does every hyperbolic triangle have a circumscribed circle? Solution: (a) Similarly to Euclidean/spherical cases, an angle bisector is a locus of points on the same distance from the rays forming the angle (this is most clear if you put the vertex of the angle at the centre of the Poincare disc model). So, the intersection point of two angle bisectors lies on the same distance from all three sides of the triangle, which implies that it actually lies on the third angle bisector. (The intersection point does exists since the two ends of one angle bisector -say AA1 -lie on two different sides of the other angle: A ∈ BA, A1 ∈ BC , so they are separated by the angle bisector BB1 ). (b) Blowing the small circle centred at the point of intersection of angle bisectors, we obtain at some point a circle tangent to all three sides. (c) The vertices of the hyperbolic triangle (in any of two Poincar´e models) are represented by vertices of Euclidean triangle. Consider a Euclidean circle γ passing through the vertices of this Euclidean triangle (it does exist by E15). If γ lies entirely inside the hyperbolic plane (i.e. in the disc or in the upper half-plane) then it represents some hyperbolic circle passing through the given points. However, the circle γ may intersect the boundary of hyperbolic plane. Then it does not represent any hyperbolic circle. Moreover, in the latter case no hyperbolic circle passes through the given points, as each hyperbolic circle is represented by some Euclidean circle.

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